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Math Help - Max/min on a continuous interval.

  1. #1
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    Max/min on a continuous interval.

    Suppose that f(x) is continuous and > 0 on I = \mathbb{R}, and \displaystyle\lim_{x\to \pm\infty}f(x) = 0.

    (a) Prove f(x) has no minimum on I.
    (b) Prove f(x) has a maximum on I (note that I is not compact).
    (c) Prove (b) under weaker hypothesis than positivity on all of I.
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  2. #2
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    Quote Originally Posted by cgiulz View Post
    Suppose that f(x) is continuous and > 0 on I = \mathbb{R}, and \displaystyle\lim_{x\to \pm\infty}f(x) = 0.

    (a) Prove f(x) has no minimum on I.
    (b) Prove f(x) has a maximum on I (note that I is not compact).
    (c) Prove (b) under weaker hypothesis than positivity on all of I.
    For (a) I'm assuming you mean a global minimum (otherwise this is not true) and it's obvious since that minimum would have to be 0 contradicting the fact that f>0.

    For (c) we're going to assume f \geq 0 for some x \in \mathbb{R}. Take an interval of the form [-K,K] and take the maximum of f there say f(x_0)=y_0>0 (if it's zero we'll deal with it later). Now since f(x) \rightarrow 0 when x \rightarrow \pm \infty we have two constants K_1,K_2>0 such that \vert f(x) \vert < y_0 whenever x>K_1 or x<-K_2 take m= \max \{K_1,K_2 \} and so we have that f attains a maximum y in [-m,m] with y \geq y_0 and since out of this interval the function never exceeds y_0, y is a global maximum.

    If y_0 were 0 and f \leq 0 in I we're finished. If it's f>0 on an unbounded interval apply a similar argument to the above, if f>0 on a bounded interval it's clear it has a maximum. And this covers all cases.
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  3. #3
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    Can anyone show me where to start in order to prove the existence of a max by positivity?

    Thanks.

    [Forget it, I realized I can just pull it out of part (c)]
    Last edited by cgiulz; October 25th 2009 at 07:17 PM.
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